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ANEK [815]
3 years ago
5

A system being studied in a 3.0-liter reactor was charged with 0.150 moles of H2, 0.150 moles of X2, and 0.600 moles of HX. A ca

talyst was introduced using a remote unit. Is the reaction at equilibrium
Chemistry
1 answer:
ycow [4]3 years ago
8 0

Answer:

The reaction is not in equilibrium and it proceed in reverse direction to reach the equilibruim condition

Explanation:

We have

H₂(g) + x₂(g)  ⇆  2HX (g)

Kc = 2.4 x 10 ⁻³ at 25°C

K_c = \frac{[HX]^2}{[H_2][X_2]}

Now we have 3L reactor and the number of mole of species are 0.150 mole H₂, 0.150 mole X₂ and 0.600 mole HX

Therefore,

[H_2] = \frac{0.150}{3} \\= 0.05M

[X_2] = \frac{0.150}{3} \\= 0.05M

[HX] = \frac{0.600}{3} \\= 0.2M

Therefore,

Q_c = \frac{[HX]^2}{[H_2][X_2]}

Q_c = \frac{[0.200]^2}{[0.05][0.05]}\\\\=16

Now, we calculate the free energy change for the reaction

ΔrG = RT In Q/K

= (8.314) \times (298) \times In (\frac{16}{2.4\times 10^-^3} )\\\\=+21814.7 J/mol

As free energy is +ve so the reaction is not in equilibruim and proceed in reverse reaction to reach tje equilibruim condition

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